| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Standard +0.3 This is a standard M1 projectile motion question with straightforward application of SUVAT equations. Parts (a)-(b) involve routine calculations with given initial conditions. Parts (c)-(d) test understanding that maximum speed occurs at lowest point, requiring simple application of v² = u² + 2as. All steps are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(-0.7 = 4t - \frac{1}{2}(9.8)t^2\) | M1 A1 | Using \(s = ut + \frac{1}{2}at^2\) vertically |
| \(4.9t^2 - 4t - 0.7 = 0\) | M1 | Rearranging |
| \(t = \frac{4 \pm \sqrt{16 + 13.72}}{9.8} \approx 0.952\text{ s}\) | A1 A1 | Positive root only |
| (b) \(x = 4\sqrt{3} \times 0.952 \approx 6.59\text{ m}\) | M1 A1 | |
| (c) \(h = 1.2\text{ m}\) (at maximum height, vertical velocity = 0) | B1 | |
| (d) Maximum speed = horizontal component \(= 4\sqrt{3} \approx 6.93\text{ ms}^{-1}\) | M1 A1 A1 A1 |
## Question 6:
**Setup:** Initial velocity $8\text{ ms}^{-1}$ at $30°$, so $u_x = 8\cos30° = 4\sqrt{3}$, $u_y = 8\sin30° = 4$. Launch height $1.2\text{ m}$, catch height $0.5\text{ m}$, so vertical displacement $= -0.7\text{ m}$
**(a)** $-0.7 = 4t - \frac{1}{2}(9.8)t^2$ | M1 A1 | Using $s = ut + \frac{1}{2}at^2$ vertically |
$4.9t^2 - 4t - 0.7 = 0$ | M1 | Rearranging |
$t = \frac{4 \pm \sqrt{16 + 13.72}}{9.8} \approx 0.952\text{ s}$ | A1 A1 | Positive root only |
**(b)** $x = 4\sqrt{3} \times 0.952 \approx 6.59\text{ m}$ | M1 A1 | |
**(c)** $h = 1.2\text{ m}$ (at maximum height, vertical velocity = 0) | B1 | |
**(d)** Maximum speed = horizontal component $= 4\sqrt{3} \approx 6.93\text{ ms}^{-1}$ | M1 A1 A1 A1 | |
---6 Emma is in a park with her dog, Roxy. Emma throws a ball and Roxy catches it in her mouth. The ground in the park is horizontal. Emma throws the ball from a point at a height of 1.2 metres above the ground and Roxy catches the ball when it is at a height of 0.5 metres above the ground. Emma throws the ball with an initial velocity of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ above the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Find the time that the ball takes to travel from Emma's hand to Roxy's mouth.
\item Find the horizontal distance travelled by the ball during its flight.
\item During the flight, the speed of the ball is a maximum when it is at a height of $h$ metres above the ground. Write down the value of $h$.
\item Find the maximum speed of the ball during its flight.\\[0pt]
[4 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{01338c87-302c-420f-a473-39b0796ccaed-14_1566_1707_1137_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2015 Q6 [12]}}